Determine the equation of the asymptote for the function:
h(x) = 7 - sec (6theta -pi over 2

Asymptotes for secant will be where cosine is 0

cos (6x - π/2) = 0
6x - π/2 = π/2
6x = π
x = π/6 → This is the first positive asymptote

The period of f(x) = sec x is 2π.
So the period of h(x) = 2π/6 = π/3
There are 2 asymptotes in each period of the secant function, so h(x) has an asymptote every π/6.

So the asymptotes are at 
where k is an integer

256825) wjjikbhjkxk$: 882&